That's the title that Julie J. Rehnmayer used for her Science News Online report on recent answers to the question: when an inelastic rectangle (for example, a strip of paper) is twisted into a Möbius band in 3-dimensional space, what exactly is the resulting shape?

An rectangle with side ratio to 1 can be folded into a Möbius Band by reassembling it as a trapezoid (a), folding along the blue dotted line (b), and then folding along the green. The last fold (c) brings into congruence, with proper orientation, the sides that are to be identified. This configuration is the limiting case of the embeddings studied by Starostin and Van der Heijden.

When the side ratio is to 1, the strip can be folded into a configuration that respects the edge identification. For narrower strips the band assumes a "characteristic shape" minimizing the total bending energy; the exact determination of this shape has been an oustanding problem at least since 1930. Evgueni Starostin and Gert van der Heijden (University College London) recently nailed down the solution using "the invariant variational bicomplex formalism" and numerical methods. (The variational bicomplex is, according to Ian Anderson, a double complex of differential forms defined on the infinite jet bundle of any fibered manifold π : E --> M.) They report: "Solutions for increasing width show the formation of creases bounding nearly flat triangular regions..." .

Three of six characteristic shapes for length = 2π and various widths shown in Srarostin and Van der Heijden's article: width 0.2 (b), 0.8 (d) and 1.5 (f). "The colouring changes according to the local bending energy density, from violet for regions of low bending to red for regions of high bending." Image from Nature Materials Online (posted July 15, 2007), used with permission.

Geometry and the Imagination

Bill Thurston (in Princeton, March 1990)

The 5-day conference with this title, held at Princeton on June 7-11 in honor of Bill Thurston's 60th birthday, was surveyed by Barry Cipra in the July 6 2007 Science. Cipra's 2-page spread covers four of the presentations.

The smallest hyperbolic manifold. In hyperbolic geometry, similar triangles must have the same area, and each hyperbolic manifold has its own specific volume. In the 1970s, Cipra tells us, "Thurston ... proved a surprising property of hyperbolic manifolds. Given any infinite collection of such manifolds, one member of the collection will be of smallest volume." In particular, one hyperbolic 3-manifold must have the smallest volume of all. A candidate was discovered shortly thereafter, by Jeff Weeks. The "Weeks manifold" remained for a long time the smallest hyperbolic 3-manifold known; only this year did David Gabai, Robert Meyerhoff and Peter Milley prove that there can be no smaller. Their work was posted on arXiv May 30, 2007.

Infinite trajectories in outer billiards. Outer billiards was devised as a simple analogue of planetary motion. "An object starting at point x0 outside some convex figure zips along a straight line just touching the figure to a new point x1 at the same distance from the point of contact. It then repeats this over and over, thereby orbiting the figure in, say, a clockwise fashion." (Cipra). Are all such orbits bounded, or for some figure and some x0 could the xi wind up arbitrarily far away? The question had been open since the 1950s, but a set of unbounded examples was recently discovered by Richard Schwartz. The convex body he uses is the kite from Penrose tilings, and he exhibits "larger and larger clouds of smaller and smaller regions" converging to "a set of points from which the trajectories are unbounded." Details at Rich's website.

Crossing number of the sum of two knots. It is known that knots can't cancel. But how about partial simplification? "... if two knots are strung together to form one larger, more complicated knot, can the new knot be redrawn with fewer crossings than the original two knots combined?" Cipra quotes Colin Adams: "This problem has been out there forever." Some recent progress towards proving that the minimal crossing number c(K1 # K2) of the knot sum is the sum c(K1) + c(K2) of those of the addends was reported by Mark Lakenby, who showed that c(K1 # K2) is at least (1/281)[c(K1) + c(K2)]. Cipra: "The basic idea is to think of each knot as enclosed in a spherical bubble and then carefully analyze what must happen to the bubbles if the knot sum is twisted into a new shape with fewer crossings." He remarks, "To prove the full conjecture, mathematicians will need to whittle this number [281] all the way down to one."

Update on the Poincaré conjecture. "Pricey Proof Keeps Gaining Support" is Cipra's heading for his report on John Morgan's overview of Perelman's proof. "After poring over Perelman's papers for 4 years, topologists are confident of the result. ... Much of the confidence derives from alternative proofs researchers have devised in the wake of Perelman's work." Cipra quotes Thurston at the conference banquet: "I never doubted it would be proved. It's really wonderful to see the community ownership of this mathematics."

Homer Simpson: "Mmm....π"

Nature's Michael Hopkin interviewed The Simpsons' executive producer, Al Jean, for a News Feature item (July 27, 2007). Jean, it turns out, was an undergraduate math major at Harvard and has kept some of the ethos in his new job. "I look at comedy writing mathematically, it's sort of like a proof in which you're trying to find the ideal punchline for a setup, and when you get it it's a very elegant feeling." He and his fellow writers enjoy slipping obscure mathematical references into the Simpsons' storyboards. Contestants in a quiz will have to choose between various numbers "and each of the options is a different mathematical irregularity -- one's a perfect number, one's a sum of four squares. They're all in the thousands and they're numbers that nobody except a mathematician would, at face value, recognize as anything unusual, but if you're really sharp you'll pick it up." The closest he comes to what passes for humor at tea in math departments is the witticism referred to in the title: "...we did an episode where Apu was a witness in a courtroom and the lawyer asked if he had a good memory. He said, yes I do, I've memorized pi to one million decimal places, and Homer said "mmm... pi" and started drooling."